Question

describe the effects the outlier has on all the measures of central tendency.

1. 3, 2, 6, 4, 3, 5, 16

with outlier
mean:
median:
mode:
range:
stdevp:
without outlier
mean:
median:
mode:
range:
stdevp:
formula for bounds
lower boundary: q1 - 1.5(iqr)
upper boundary: q3 + 1.5(iqr)
lower bound:
upper bound:

Explanation:

Step1: Identify the outlier

The data set is \(3, 2, 6, 4, 3, 5, 16\). The outlier here is \(16\) as it is much larger than the other values.

Step2: Calculate with outlier (all values: \(2, 3, 3, 4, 5, 6, 16\))

Mean:

Sum of values \(= 2 + 3 + 3 + 4 + 5 + 6 + 16 = 39\)
Number of values \(n = 7\)
Mean \(=\frac{39}{7}\approx 5.57\)

Median:

Arrange in order: \(2, 3, 3, 4, 5, 6, 16\). Middle value (4th term) is \(4\).

Mode:

The most frequent value is \(3\) (appears twice).

Range:

Maximum - Minimum \(= 16 - 2 = 14\)

Stdevp (Population Standard Deviation):

First, find the mean \(\mu=\frac{39}{7}\approx 5.57\)
Sum of squared deviations: \((2 - 5.57)^2+(3 - 5.57)^2+(3 - 5.57)^2+(4 - 5.57)^2+(5 - 5.57)^2+(6 - 5.57)^2+(16 - 5.57)^2\)
\(= (-3.57)^2+(-2.57)^2+(-2.57)^2+(-1.57)^2+(-0.57)^2+(0.43)^2+(10.43)^2\)
\(= 12.7449+6.6049+6.6049+2.4649+0.3249+0.1849+108.7849\)
\(= 137.7143\)
Stdevp \(=\sqrt{\frac{137.7143}{7}}\approx\sqrt{19.6735}\approx 4.435\)

Step3: Calculate without outlier (values: \(2, 3, 3, 4, 5, 6\))

Mean:

Sum of values \(= 2 + 3 + 3 + 4 + 5 + 6 = 23\)
Number of values \(n = 6\)
Mean \(=\frac{23}{6}\approx 3.83\)

Median:

Arrange in order: \(2, 3, 3, 4, 5, 6\). Middle values are \(3\) and \(4\). Median \(=\frac{3 + 4}{2}= 3.5\)

Mode:

The most frequent value is \(3\) (appears twice).

Range:

Maximum - Minimum \(= 6 - 2 = 4\)

Stdevp (Population Standard Deviation):

First, find the mean \(\mu=\frac{23}{6}\approx 3.83\)
Sum of squared deviations: \((2 - 3.83)^2+(3 - 3.83)^2+(3 - 3.83)^2+(4 - 3.83)^2+(5 - 3.83)^2+(6 - 3.83)^2\)
\(= (-1.83)^2+(-0.83)^2+(-0.83)^2+(0.17)^2+(1.17)^2+(2.17)^2\)
\(= 3.3489+0.6889+0.6889+0.0289+1.3689+4.7089\)
\(= 10.8334\)
Stdevp \(=\sqrt{\frac{10.8334}{6}}\approx\sqrt{1.8056}\approx 1.344\)

Step4: Bounds (to confirm outlier)

First, find \(Q1\) and \(Q3\) for the original data (with outlier)
Data: \(2, 3, 3, 4, 5, 6, 16\)
Median (Q2) is \(4\) (4th term)
Lower half: \(2, 3, 3\) (Q1 is the median of lower half, so \(Q1 = 3\))
Upper half: \(5, 6, 16\) (Q3 is the median of upper half, so \(Q3 = 6\))
Lower Boundary: \(Q1 - 1.5(IQR)\), \(IQR=Q3 - Q1 = 6 - 3 = 3\), so Lower Bound \(= 3-1.5\times3=3 - 4.5=- 1.5\)
Upper Boundary: \(Q3 + 1.5(IQR)=6 + 1.5\times3=6 + 4.5 = 10.5\)
Since \(16>10.5\), it is an outlier.

Answer:

With Outlier:

• Mean: \(\boldsymbol{\approx 5.57}\)
• Median: \(\boldsymbol{4}\)
• Mode: \(\boldsymbol{3}\)
• Range: \(\boldsymbol{14}\)
• Stdevp: \(\boldsymbol{\approx 4.435}\)

Without Outlier:

• Mean: \(\boldsymbol{\approx 3.83}\)
• Median: \(\boldsymbol{3.5}\)
• Mode: \(\boldsymbol{3}\)
• Range: \(\boldsymbol{4}\)
• Stdevp: \(\boldsymbol{\approx 1.344}\)

Bounds:

• Lower Bound: \(\boldsymbol{-1.5}\)
• Upper Bound: \(\boldsymbol{10.5}\)